tea.mathoverflow.net - Discussion Feed (Doob's inequality)Mon, 19 Aug 2019 23:56:37 -0700
http://mathoverflow.tqft.net/
Lussumo Vanilla 1.1.9 & Feed Publisher
George Lowther comments on "Doob's inequality" (20823)
http://mathoverflow.tqft.net/discussion/1487/doobs-inequality/?Focus=20823#Comment_20823
http://mathoverflow.tqft.net/discussion/1487/doobs-inequality/?Focus=20823#Comment_20823Fri, 14 Dec 2012 14:31:35 -0800George Lowther
Yes, I agree that this is not really suitable for mathoverflow, and is more likely to get an answer at math.stackexchange. Also, the inequality mentioned is a special case of the following question (and answer) already asked on math.SE http://math.stackexchange.com/q/88371/1321, and the technique is to use time-change.
]]>
HJRW comments on "Doob's inequality" (20821)
http://mathoverflow.tqft.net/discussion/1487/doobs-inequality/?Focus=20821#Comment_20821
http://mathoverflow.tqft.net/discussion/1487/doobs-inequality/?Focus=20821#Comment_20821Fri, 14 Dec 2012 13:55:56 -0800HJRW
To clarify, I think MemT is asking if this question could be reopened. As s/he has explained that s/he's revising for an exam, I have explained that this is not on-topic for MO and recommended math.stackexchange.com.
]]>
MemT comments on "Doob's inequality" (20818)
http://mathoverflow.tqft.net/discussion/1487/doobs-inequality/?Focus=20818#Comment_20818
http://mathoverflow.tqft.net/discussion/1487/doobs-inequality/?Focus=20818#Comment_20818Fri, 14 Dec 2012 13:03:31 -0800MemT
$$\mathcal{P}\left[\sup_{0\leq t\leq T}|I_t|>\lambda\right]\leq \exp\left(-\frac{\lambda^2}{2M^2T}\right).$$

First I tried by defining $Y_t^{\alpha}=\exp\left(\alpha I_t-\frac{1}{2}\int_{0}^t f^{2}(s)ds\right)$, where $\alpha\in \mathbb{R}$ to get an upper bound. But I need to know how to show that $Y_t^{\alpha}$ is a mgale. Thank ]]>